Exercise 5.3 (Revised) - Chapter 5 - Continuity & Differentiability - Ncert Solutions class 12 - Maths

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Chapter 5: Continuity & Differentiability NCERT Solutions Class 12 Maths

Ex 5.3 Question 1 :

Find $\frac{d y}{d x}$ :
$
2 x+3 y=\sin x
$

Answer :
The given relationship is $2 x+3 y=sin x$
Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}(2 x+3 y)=\frac{d}{d x}(\sin x) \\
& \Rightarrow \frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\cos x \\
& \Rightarrow 2+3 \frac{d y}{d x}=\cos x \\
& \Rightarrow 3 \frac{d y}{d x}=\cos x-2 \\
& \therefore \frac{d y}{d x}=\frac{\cos x-2}{3}
\end{aligned}
$

Ex 5.3 Question 2:

Find $\frac{d y}{d x}$ :2 x+3 y=sin y

Answer :
The given relationship is $2 x+3 y=\sin y$
Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\frac{d}{d x}(\sin y) \\
& \Rightarrow 2+3 \frac{d y}{d x}-\cos y \frac{d y}{d x} \quad \text { [By using chain rule] } \\
& \Rightarrow 2=(\cos y-3) \frac{d y}{d x} \\
& \therefore \frac{d y}{d x}=\frac{2}{\cos y-3}
\end{aligned}
$
[By using chain rule]

Ex 5.3 Question 3:

Find $\frac{d y}{d x}$ :
$
a x+b y^2=\cos y
$

Answer:
The given relationship is $a x+b y^2-\cos y$

Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}(a x)+\frac{d}{d x}\left(b y^2\right)=\frac{d}{d x}(\cos y) \\
& \Rightarrow a+b \frac{d}{d x}\left(y^2\right)=\frac{d}{d x}(\cos y)
\end{aligned}
$

Using chain rule, we obtain $\frac{d}{d x}\left(y^2\right)=2 y \frac{d y}{d x}$ and $\frac{d}{d x}(\cos y)=-\sin y \frac{d y}{d x}$
From (1) and (2), we obtain
$
\begin{aligned}
& a+b \times 2 y \frac{d y}{d x}=-\sin y \frac{d y}{d x} \\
& \Rightarrow(2 b y+\sin y) \frac{d y}{d x}--a \\
& \therefore \frac{d y}{d x}=\frac{-a}{2 b y+\sin y}
\end{aligned}
$

Ex 5.3 Question 4:

Find $\frac{d y}{d x}$ :
$
x y+y^2=\tan x+y
$

Answer :
The given relationship is $x y+y^2=\tan x+y$
Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}\left(x y+y^2\right)=\frac{d}{d x}(\tan x+y) \\
& \Rightarrow \frac{d}{d x}(x y)+\frac{d}{d x}\left(y^2\right)=\frac{d}{d x}(\tan x)+\frac{d y}{d x} \\
& \Rightarrow\left[y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}\right]+2 y \frac{d y}{d x}=\sec ^2 x+\frac{d y}{d x} \\
& \Rightarrow y \cdot 1+x \cdot \frac{d y}{d x}+2 y \frac{d y}{d x}=\sec ^2 x+\frac{d y}{d x} \\
& \Rightarrow(x+2 y-1) \frac{d y}{d x}=\sec ^2 x-y \\
& \therefore \frac{d y}{d x}=\frac{\sec ^2 x-y}{(x+2 y \quad 1)}
\end{aligned}
$
[Using product rule and chain rule]

Ex 5.3 Question 5:

Find $\frac{d y}{d x}$ :
$
x^2+x y+y^2=100
$

Answer :
The given relationship is $x^2+x y+y^2=100$
Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}\left(x^2+x y+y^2\right)=\frac{d}{d x}(100) \\
& \Rightarrow \frac{d}{d x}\left(x^2\right)+\frac{d^{\prime}}{d x}(x y)+\frac{d}{d x}\left(y^2\right)=0 \text { [Derivative of constant function is } 0 \text { ] } \\
& \Rightarrow 2 x+\left[y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}\right]+2 y \frac{d y}{d x}=0 \quad \text { [Using product rule and chain rule] } \\
& \Rightarrow 2 x+y \cdot 1+x \cdot \frac{d y}{d x}+2 y \frac{d y}{d x}=0 \\
& \Rightarrow 2 x+y+(x+2 y) \frac{d y}{d x}=0 \\
& \therefore \frac{d y}{d x}=-\frac{2 x+y}{x+2 y}
\end{aligned}
$

Ex 5.3 Question 6:

Find $\frac{d y}{d x}$ :
$
x^3+x^2 y+x y^2+y^3=81
$

Answer:
The given relationship is $x^3+x^2 y+x y^2+y^3=81$
Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}\left(x^3+x^2 y+x y^2+y^3\right)=\frac{d}{d x}(81) \\
& \Rightarrow \frac{d}{d x}\left(x^3\right)+\frac{d}{d x}\left(x^2 y\right)+\frac{d}{d x}\left(x y^2\right)+\frac{d}{d x}\left(y^3\right)=0 \\
& \Rightarrow 3 x^2+\left[y \frac{d}{d x}\left(x^2\right)+x^2 \frac{d y}{d x}\right]+\left[y^2 \frac{d}{d x}(x)+x \frac{d}{d x}\left(y^2\right)\right]+3 y^2 \frac{d y}{d x}=0 \\
& \Rightarrow 3 x^2+\left[y \cdot 2 x+x^2 \frac{d y}{d x}\right]+\left[y^2 \cdot 1+x \cdot 2 y \cdot \frac{d y}{d x}\right]+3 y^2 \frac{d y}{d x}-0 \\
& \Rightarrow\left(x^2+2 x y+3 y^2\right) \frac{d y}{d x}+\left(3 x^2+2 x y+y^2\right)=0 \\
& \therefore \frac{d y}{d x}=\frac{-\left(3 x^2+2 x y+y^2\right)}{\left(x^2+2 x y+3 y^2\right)}
\end{aligned}
$

Ex 5.3 Question 7:

Find $\frac{d y}{d x}$ :
$
\sin ^2 y+\cos x y=\pi
$

Answer :
The given relationship is $\sin ^2 y+\cos x y=\pi$
Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}\left(\sin ^2 y+\cos x y\right)=\frac{d}{d x}(\pi) \\
& \Rightarrow \frac{d}{d x}\left(\sin ^2 y\right)+\frac{d}{d x}(\cos x y)=0
\end{aligned}
$

Using chain rule, we obtain
$
\begin{aligned}
\frac{d}{d x}\left(\sin ^2 y\right) & =2 \sin y \frac{d}{d x}(\sin y)=2 \sin y \cos y \frac{d y}{d x} \\
\frac{d}{d x}(\cos x y) & =-\sin x y \frac{d}{d x}(x y)=-\sin x y\left[y \frac{d}{d x}(x)+x \frac{d y}{d x}\right] \\
& =-\sin x y\left[y \cdot 1+x \frac{d y}{d x}\right]=-y \sin x y-x \sin x y \frac{d y}{d x}
\end{aligned}
$

From (1), (2), and (3), we obtain
$
\begin{aligned}
& 2 \sin y \cos y \frac{d y}{d x}-y \sin x y-x \sin x y \frac{d y}{d x}=0 \\
& \Rightarrow(2 \sin y \cos y-x \sin x y) \frac{d y}{d x}=y \sin x y \\
& \Rightarrow(\sin 2 y-x \sin x y) \frac{d y}{d x}=y \sin x y \\
& \therefore \frac{d y}{d x}=\frac{y \sin x y}{\sin 2 y-x \sin x y}
\end{aligned}
$

Ex 5.3 Question 8:

Find $\frac{d y}{d x}$ :
$
\sin ^2 x+\cos ^2 y=1
$

Answer:
The given relationship is $\sin ^2 x+\cos ^2 y=1$

Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}\left(\sin ^2 x+\cos ^2 y\right)=\frac{d}{d x}(1) \\
& \Rightarrow \frac{d}{d x}\left(\sin ^2 x\right)+\frac{d}{d x}\left(\cos ^2 y\right)=0 \\
& \Rightarrow 2 \sin x \cdot \frac{d}{d x}(\sin x)+2 \cos y \cdot \frac{d}{d x}(\cos y)=0 \\
& \Rightarrow 2 \sin x \cos x+2 \cos y(-\sin y) \cdot \frac{d y}{d x}=0 \\
& \Rightarrow \sin 2 x-\sin 2 y \frac{d y}{d x}=0 \\
& \therefore \frac{d y}{d x}=\frac{\sin 2 x}{\sin 2 y}
\end{aligned}
$

Ex 5.3 Question 9:

Find $\frac{d y}{d x}$ :
$
y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)
$

Answer:
The given relationship is $y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$
$
\begin{aligned}
& y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \\
& \Rightarrow \sin y=\frac{2 x}{1+x^2}
\end{aligned}
$

Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}(\sin y)=\frac{d}{d x}\left(\frac{2 x}{1+x^2}\right) \\
& \Rightarrow \cos y \frac{d y}{d x}=\frac{d}{d x}\left(\frac{2 x}{1+x^2}\right)
\end{aligned}
$

The function, $\frac{2 x}{1+x^2}$, is of the form of $\frac{u}{v}$.
Therefore, by quotient rule, we obtain

$
\begin{aligned}
& \frac{d}{d x}(\sin y)=\frac{d}{d x}\left(\frac{2 x}{1+x^2}\right) \\
& \Rightarrow \cos y \frac{d y}{d x}=\frac{d}{d x}\left(\frac{2 x}{1+x^2}\right)
\end{aligned}
$

The function, $\frac{2 x}{1+x^2}$, is of the form of $\frac{u}{v}$.
Therefore, by quotient rule, we obtain
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{2 x}{1+x^2}\right)=\frac{\left(1+x^2\right) \cdot \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}\left(1+x^2\right)}{\left(1+x^2\right)^2} \\
& =\frac{\left(1+x^2\right) \cdot 2-2 x \cdot[0+2 x]}{\left(1+x^2\right)^2}=\frac{2+2 x^2-4 x^2}{\left(1+x^2\right)^2}=\frac{2\left(1-x^2\right)}{\left(1+x^2\right)^2}
\end{aligned}
$

Also, $\sin y=\frac{2 x}{1+x^2}$
$
\begin{aligned}
\Rightarrow \cos y & =\sqrt{1-\sin ^2 y}=\sqrt{1-\left(\frac{2 x}{1+x^2}\right)^2}=\sqrt{\frac{\left(1+x^2\right)^2-4 x^2}{\left(1+x^2\right)^2}} \\
& =\sqrt{\frac{\left(1-x^2\right)^2}{\left(1+x^2\right)^2}}=\frac{1-x^2}{1+x^2}
\end{aligned}
$

From (1), (2), and (3), we obtain
$
\frac{1-x^2}{1+x^2} \times \frac{d y}{d x}=\frac{2\left(1-x^2\right)}{\left(1+x^2\right)^2}
$

$\Rightarrow \frac{d y}{d x}=\frac{2}{1+x^2}$.

Ex 5.3 Question 10:

Find $\frac{d y}{d x}$ :
$
y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right),-\frac{1}{\sqrt{3}} $

Answer :
The given relationship is $y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)$
$
\begin{aligned}
& y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right) \\
& \Rightarrow \tan y=\frac{3 x-x^3}{1-3 x^2}
\end{aligned}
$

It is known that, $\tan y=\frac{3 \tan \frac{y}{3}-\tan ^3 \frac{y}{3}}{1-3 \tan ^2 \frac{y}{3}}$
Comparing equations (1) and (2), we obtain
$
x=\tan \frac{y}{3}
$

Differentiating this relationship with respect to $x$, we obtain
$
\frac{d}{d x}(x)=\frac{d}{d x}\left(\tan \frac{y}{3}\right)
$

$\begin{aligned}
& \Rightarrow 1=\sec ^2 \frac{y}{3} \cdot \frac{d}{d x}\left(\frac{y}{3}\right) \\
& \Rightarrow 1=\sec ^2 \frac{y}{3} \cdot \frac{1}{3} \cdot \frac{d y}{d x} \\
& \Rightarrow \frac{d y}{d x}=\frac{3}{\sec ^2 \frac{y}{3}}=\frac{3}{1+\tan ^2 \frac{y}{3}} \\
& \therefore \frac{d y}{d x}-\frac{3}{1+x^2}
\end{aligned}$

Ex 5.3 Question 11:

Find $\frac{d y}{d x}$ :
$
y=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 $

Answer:
The given relationship is,
$
\begin{aligned}
& y=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\
& \Rightarrow \cos y=\frac{1-x^2}{1+x^2} \\
& \Rightarrow \frac{1-\tan ^2 \frac{y}{2}}{1+\tan ^2 \frac{y}{2}}=\frac{1-x^2}{1+x^2}
\end{aligned}
$

On comparing L.H.S. and R.H.S. of the above relationship, we obtain
$
\tan \frac{y}{2}=x
$

Differentiating this relationship with respect to $x$, we obtain

Ex 5.3 Question 12:

Find $\frac{d y}{d x}$ :
$
y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 $

Answer :
The given relationship is $y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
$
\begin{aligned}
& y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\
& \Rightarrow \sin y=\frac{1-x^2}{1+x^2}
\end{aligned}
$

Differentiating this relationship with respect to $x$, we obtain
$
\frac{d}{d x}(\sin y)=\frac{d}{d x}\left(\frac{1-x^2}{1+x^2}\right)
$
using chain rule, we obtain
$
\frac{d}{d x}(\sin y)=\cos y \cdot \frac{d y}{d x}
$

$\begin{aligned}
\cos y & =\sqrt{1-\sin ^2 y}=\sqrt{1-\left(\frac{1-x^2}{1+x^2}\right)^2} \\
& =\sqrt{\frac{\left(1+x^2\right)^2-\left(1-x^2\right)^2}{\left(1+x^2\right)^2}}=\sqrt{\frac{4 x^2}{\left(1+x^2\right)^2}}=\frac{2 x}{1+x^2}
\end{aligned}$

$
\therefore \frac{d}{d x}(\sin y)=\frac{2 x}{1+x^2} \frac{d y}{d x}
$
$
\frac{d}{d x}\left(\frac{1-x^2}{1+x^2}\right)=\frac{\left(1+x^2\right) \cdot\left(1-x^2\right)^{\prime}-\left(1-x^2\right) \cdot\left(1+x^2\right)^{\prime}}{\left(1+x^2\right)^2}
$
[Using quotient rule]

$
\begin{aligned}
& =\frac{\left(1+x^2\right)(-2 x)-\left(1-x^2\right) \cdot(2 x)}{\left(1+x^2\right)^2} \\
& =\frac{-2 x-2 x^3-2 x+2 x^3}{\left(1+x^2\right)^2} \\
& =\frac{-4 x}{\left(1+x^2\right)^2}
\end{aligned}
$

from (1), (2), and (3), we obtain
$
\begin{aligned}
& \frac{2 x}{1+x^2} \frac{d y}{d x}-\frac{-4 x}{\left(1+x^2\right)^2} \\
& \Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^2}
\end{aligned}
$

Alternate method
$
\begin{aligned}
& y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\
& \Rightarrow \sin y=\frac{1-x^2}{1+x^2} \\
& \Rightarrow\left(1+x^2\right) \sin y=1-x^2 \\
& \Rightarrow(1+\sin y) x^2=1-\sin y \\
& \Rightarrow x^2=\frac{1-\sin y}{1+\sin y} \\
& \Rightarrow x^2=\frac{\left(\cos \frac{y}{2}-\sin \frac{y}{2}\right)^2}{\left(\cos \frac{y}{2}+\sin \frac{y}{2}\right)^2} \\
& \Rightarrow x=\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}} \\
& \Rightarrow x=\frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}} \\
& \Rightarrow x=\tan \left(\frac{\pi}{4}-\frac{y}{2}\right)
\end{aligned}
$

Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}(x)=\frac{d}{d x} \cdot\left[\tan \left(\frac{\pi}{4}-\frac{y}{2}\right)\right] \\
& \Rightarrow 1=\sec ^2\left(\frac{\pi}{4}-\frac{y}{2}\right) \cdot \frac{d}{d x}\left(\frac{\pi}{4}-\frac{y}{2}\right) \\
& \Rightarrow 1=\left[1+\tan ^2\left(\begin{array}{ll}
\pi & y \\
4 & 2
\end{array}\right)\right] \cdot\left(\begin{array}{cc}
1 & d y \\
2 & d x
\end{array}\right) \\
& \Rightarrow 1=\left(1+x^2\right)\left(-\frac{1}{2} \frac{d y}{d x}\right) \\
& \Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^2}
\end{aligned}
$

Ex 5.3 Question 14:

Find $\frac{d y}{d x}$ :
$
y=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right),-\frac{1}{\sqrt{2}} $

Answer :
The given relationship is $y=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$
$
\begin{aligned}
& y=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right) \\
& \Rightarrow \sin y=2 x \sqrt{1-x^2}
\end{aligned}
$

Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \cos y \frac{d y}{d x}=2\left[x \frac{d}{d x}\left(\sqrt{1-x^2}\right)+\sqrt{1-x^2} \frac{d x}{d x}\right] \\
& \Rightarrow \sqrt{1-\sin ^2 y} \frac{d y}{d x}=2\left[\frac{x}{2} \cdot \frac{-2 x}{\sqrt{1-x^2}}+\sqrt{1-x^2}\right] \\
& \Rightarrow \sqrt{1-\left(2 x \sqrt{1-x^2}\right)^2} \frac{d y}{d x}=2\left[\frac{-x^2+1-x^2}{\sqrt{1-x^2}}\right] \\
& \Rightarrow \sqrt{1-4 x^2\left(1-x^2\right)} \frac{d y}{d x}=2\left[\frac{1-2 x^2}{\sqrt{1-x^2}}\right] \\
& \Rightarrow \sqrt{\left(1-2 x^2\right)^2} \frac{d y}{d x}=2\left[\frac{1-2 x^2}{\sqrt{1-x^2}}\right] \\
& \Rightarrow\left(1-2 x^2\right) \frac{d y}{d x}=2\left[\frac{1-2 x^2}{\sqrt{1-x^2}}\right] \\
& \Rightarrow \frac{d y}{d x}=\frac{2}{\sqrt{1-x^2}}
\end{aligned}
$

Ex 5.3 Question 15:

Find $\frac{d y}{d x}$ :
$
y=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right), 0 $

Answer :
The given relationship is $y=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$
$
\begin{aligned}
& y=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right) \\
& \Rightarrow \sec y=\frac{1}{2 x^2-1} \\
& \Rightarrow \cos y=2 x^2-1 \\
& \Rightarrow 2 x^2=1+\cos y \\
& \Rightarrow 2 x^2=2 \cos ^2 \frac{y}{2} \\
& \Rightarrow x=\cos \frac{y}{2}
\end{aligned}
$

Differentlating this relationship with respect to $x$, we obtain

$\begin{aligned}
& \frac{d}{d x}(x)=\frac{d}{d x}\left(\cos \frac{y}{2}\right) \\
& \Rightarrow 1=-\sin \frac{y}{2} \cdot \frac{d}{d x}\left(\frac{y}{2}\right) \\
& \Rightarrow \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2} \frac{d y}{d x} \\
& \Rightarrow \frac{d y}{d x}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-\cos ^2 \frac{y}{2}}} \\
& \Rightarrow \frac{d y}{d x}=\frac{-2}{\sqrt{1-x^2}}
\end{aligned}$